91影视

$f(x,y)=\tan 鈦�(x+y)\text{at point}P=\left(x_0,y_0\right)=(0,蟺)$

The line of tangent equation is

$f_x\left(x_0,y_0\right)\left(x鈭�x_0\right)+f_y\left(x_0,y_0\right)\left(y鈭�y_0\right)=z鈭�f\left(x_0,y_0\right)$

$f_x(0,蟺)(x鈭�0)+f_y(0,蟺)(y鈭�蟺)=z鈭�f(0,蟺)$$\left(1\right)$

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\tan 鈦�(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}(\tan 鈦�(x+y\left)\right)\frac{d}{dx}(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.{\sec }^2鈦�(x+y)鈰�1\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.{\sec }^2鈦�(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x(0,蟺)={\left.{\sec }^2鈦�(x+y)\right|}_{(0,蟺)}$

$f_x(0,蟺)={\sec }^2鈦�(0+蟺)$

$f_x(0,蟺)={\sec }^2鈦�\left(蟺\right)$

$f_x(0,蟺)=\frac{1}{{\cos }^2鈦�\left(蟺\right)}=\frac{1}{(\cos 鈦�(蟺){)}^2}$

$(鈭�\cos 鈦�(蟺)=鈭�1)$

$f_x(0,蟺)=\frac{1}{(鈭�1{)}^2}$

$f_x(0,蟺)=1$$\left(2\right)$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\tan 鈦�(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}(\tan 鈦�(x+y\left)\right)\frac{d}{dy}(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.{\sec }^2鈦�(x+y)鈰�1\right|}_{\left(x_0{y}_0\right)}$

$f_y\left(x_0,y_0\right)={\left.{\sec }^2鈦�(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y(0,蟺)={\sec }^2鈦�(0+蟺)$

$f_y(0,蟺)={\sec }^2鈦�\left(蟺\right)$

$f_y(0,蟺)=\frac{1}{{\cos }^2鈦�\left(蟺\right)}=\frac{1}{(\cos 鈦�(蟺){)}^2}$

$(鈭�\cos 鈦�(蟺)=鈭�1)$$f_y(0,蟺)=\frac{1}{(鈭�1{)}^2}$

$f_y(0,蟺)=1$$\left(3\right)$

$f\left(x_0,y_0\right)=f(0,蟺)=\tan 鈦�(0+蟺)$

$f(0,蟺)=\tan 鈦�\left(蟺\right)$

$f(0,蟺)=0$$\left(4\right)$

Substituting equation$f_x(0,蟺)=1$,$f_y(0,蟺)=1$and $f(0,蟺)=0$in $f_x(0,蟺)(x鈭�0)+f_y(0,蟺)(y鈭�蟺)=z鈭�f(0,蟺)$$1(x鈭�0)+1(y鈭�蟺)=z鈭�0$get

$x鈭�0+y鈭�蟺=z$

$x+y鈭�z=蟺$